has no integer solutions. I can do it easily for all cases except where z has a factor of 3, in which case I don't know what to do.

I am assuming the 3 in front of the z term is supposed to make this easier somehow than the same equation without it, but I'm failing to see the simplification that this allows.

Anyone know?

The equation can be rewritten as
Here, is expressed as the sum of two integers and which I claim is impossible.
I could restate the above statement in the following way:
I claim that it is impossible to break a perfect cube in two integers a and b such that as well as , both are perfect cubes.
Hence there are no solutions to the given equation.
The claim can be easily proved.
Keep Smiling
Malay

I claim that it is impossible to break a perfect cube in two integers a and b such that as well as , both are perfect cubes.
Hence there are no solutions to the given equation.
The claim can be easily proved.

Let
and
hence,
Now,
which cannot be a perfect cube for any value of k.
Hence the clain is proved and the question is solved.

It's a real pain to type out in a forum. But it's pretty simple with the information I posted if you understand basic algebraic number theory. If you tell me where you are getting stuck I could help a bit if you give it a try.

Let
and
hence,
Now,
which cannot be a perfect cube for any value of k.
Hence the clain is proved and the question is solved.

Keep Smiling

Malay

You do realize this doesn't make sense, right? And even if you did manage to show that FLT for p=3 implies is a perfect cube for integers z and k, you haven't proven that this is impossible, just stated it.

You do realize this doesn't make sense, right? And even if you did manage to show that FLT for p=3 implies is a perfect cube for integers z and k, you haven't proven that this is impossible, just stated it.

Could you show that is a perfect cube for any positive integer z and integer k?
Let
Let z be a given natural number.
I will prove that there is no integer k(k is an integer for which is a perfect cube) such that S is a perfect cube.
S is a perfect cube if and only if
that is, S is a perfect cube iff
which has no solutions.
Hence , S can never be a perfect cube.
Keep Smiling
Malay

I have just completed the cube.
You would have sometimes completed the square.
If you an expression, and you want to find in what conditions it is a perfect square, we would do some adjustments and express it as
For it to be a perfect square, and should be a perfect square.
I have done exactly the same thing, I have compared S with and equated the term with .
You can see it in this way also.
For S to be a perfect cube has to be zero for some value of k for a given z which is impossible.

You can see it in this way also.
For S to be a perfect cube has to be zero for some value of k for a given z which is impossible.

Keep Smiling
Malay

That's not necessarily true. You have stated this without showing it.

Why does that term have to be 0? For example, you have:

-8=64-72

-8 is a cube, 64 is a cube, and 72 is divisible by 3, so at first quick glance you can't rule it out, and even if you do rule it out, you haven't shown that nothing works.

You keep doing this Malay. You seem to misunderstand a fundamental idea about proofs and assumptions. I would suggest you spend some time improving that understanding before continuing with this line of attack or you will likely continue to make the same type of error over and over again.

The p=3 case of FLT does not have a simple solution (though it is simpler than most of the higher order cases). I don't know of any that don't use infinite descent.

I have just completed the cube.
You would have sometimes completed the square.
If you an expression, and you want to find in what conditions it is a perfect square, we would do some adjustments and express it as
For it to be a perfect square, and should be a perfect square.
I have done exactly the same thing, I have compared S with and equated the term with .